Question

The condition that the roots of $x^3+3p{x}^2+3qx+r=0$ are in H.P. is

• A. $2p^3-3pqr+r^2=0$
• B. $3p^3-2pqr+p^2=0$
• C. $2q^3-3pqr+r^2=0$
• D. $r^3-3pqr+2q^3=0$

Answer 1

The correct option is C $2q^3-3pqr+r^2=0$

Given that the roots of $x^3+{3px}^2+3qx+r=0$ are in H.P. $\left(1\right)$

Let $x=\frac{1}{x}$

$\Rightarrow ({\frac{1}{x})}^3+3p(\frac{1}{x}{)}^2+3q(\frac{1}{x})+r=0$

Multiplying by $x^3$ throughout; we get

$x^3({\frac{1}{x})}^3+x^{3}3p(\frac{1}{x}{)}^2+x^{3}3q(\frac{1}{x})+r{x}^3=0$

$\Rightarrow r{x}^3+3q{x}^2+3px+1=0$ $\left(2\right)$

The roots of the equation $\left(2\right)$ being reciprocal of the roots of the equation $\left(1\right)$ must be in A.P.

Let the roots of eq. $\left(2\right)$ be $\alpha -\beta ,\alpha ,\alpha +\beta$

$\therefore$ from equation $\left(2\right)$; we get

sum of the roots=$\alpha -\beta +\alpha +\alpha +\beta =-\frac{3q}{r}$

$\Rightarrow 3\alpha =-\frac{3q}{r}$

$\Rightarrow \alpha =-\frac{q}{r}$

Since $\alpha$ is a root of $\left(2\right)$

$\therefore r{\left(\frac{-q}{r}\right)}^3+3q(\frac{-q}{r}{)}^2+3p\left(\frac{-q}{r}\right)+1=0$

$\Rightarrow \frac{-q^3}{r^2}+\frac{{3q}^3}{r^2}-\frac{3pq}{r}+1=0$

$\Rightarrow 2q^3-3pqr+r^2=0$

$\therefore$ The required condition is $2q^3-3pqr+r^2=0$